How do you solve #abs(3x-4)<=x#?

3 Answers
Jan 18, 2017

Answer:

#|3x-4|<=x => x in [1,2]#

or #1<=x<=2#

Explanation:

2 cases

#3x-4# is positive

#|3x-4|>=0 => 3x-4=3x-4<=x#

or

#3x-4# is negative

#|3x-4|<0 => |3x-4|=4-3x<=x#

If #3x-4# is positive then,

#3x-4<=x#

#<=>#

#3x<=x+4# Add 4 to both sides

#<=>#

#3x-x=2x<=4# Subtract x from both sides

#<=>#

#x<=2# Divide both sides by 2

Or #x in (-oo,2]#

If #3x-4# is negative then,

#4-3x<=x#

#<=>#

#4<=4x# Add #3x# to both sides

#<=>#

#1<=x# Divide both sides by 4

Or #x in[1,oo)#

Notice that

#(-oo,2]nn[1,oo)=[1,2]#

Then

#x in [1,2]#

Jan 18, 2017

Answer:

#x in [1, 2]#

Explanation:

Here's one method...

Given:

#abs(3x-4) <= x#

Note in passing that #abs(...) >= 0# and hence #x >= 0#.

Given that #x >= 0#, we can square both sides of the inequality to get:

#9x^2-24x+16 <= x^2#

Subtract #x^2# from both sides to get:

#8x^2-24x+16 <= 0#

Divide both sides by #8# to get:

#x^2-3x+2 <= 0#

Factorise the quadratic to find:

#(x-1)(x-2) <= 0#

So this is a parabola with positive #x^2# coefficient, intersecting the #x# axis at #(1, 0)# and #(2, 0)#

Hence the solution of our inequality is:

#1 <= x <= 2#

In interval notation:

#x in [1, 2]#

Jan 18, 2017

Answer:

#x in [1, 2]#. See the segment of of the x-axisl in the Socratic graph. for this solution.

Explanation:

graph{(|3x-4|-x-y)<=0 [-5, 5, -2.5, 2.5]}

#|3x-4|<=x# is the combined inequality for

#3x-4<=x#, giving #x<=2#, when #x>=4/3# and

#-(3x-4)<=x#, giving #x>=1#, when #x<=4/3#.